Differential Equations JEE 2018: Official Solution and Analysis

What was the differential equations JEE 2018 question from JEE Advanced?

A twice differentiable function f on (0, π) satisfies the limit as t approaches x of [f(x) sin t − f(t) sin x] divided by (t − x) equals sin²x for every x in (0, π). It also gave f(π/6) = −π/12 and listed four statements to evaluate as true or false.

How do you solve the differential equations JEE 2018 question using the official method?

The official solution gives f(x) = −x sin x. This makes A false while B, C and D are true.

The given limit takes the 0/0 form as t approaches x. Apply L'Hospital's rule with respect to t to obtain

Rearrange to

The left side matches exactly the numerator in the derivative

Therefore

Integrate both sides to get

so $f(x)=(-x+C)\sin x.$

Substitute x = π/6 and the given value f(π/6) = −π/12:

Since sin(π/6) = 1/2 this simplifies to

yielding C = 0. Hence f(x) = −x sin x.

Direct computation at x = π/4 gives f(π/4) = −(π/4)(1/√2) = −π/(4√2), so statement A is false.

For statement B define g(x) = sin x − x + x³/6. Then g(0) = 0 and g'(0) = 0. Also g''(x) = x − sin x. On (0, π) we have x > sin x, so g''(x) > 0. Thus g' is strictly increasing and g'(x) > g'(0) = 0 for x > 0. Therefore g is strictly increasing from g(0) = 0, which means g(x) > 0 on (0, π). This proves sin x > x − x³/6 and confirms B is true.

Differentiate f(x) = −x sin x to obtain f'(x) = −sin x − x cos x. Then f'(π/2) = −1 < 0 while

Since f' is continuous, the intermediate value theorem guarantees a root α in (π/2, 3π/4) ⊂ (0, π), so C is true.

Next, f''(x) = −2 cos x + x sin x. At x = π/2 we have f''(π/2) = π/2 and f(π/2) = −π/2, so their sum is zero and D is true.

The correct options are therefore B, C and D.

What reasoning error leads students to mark the wrong option in differential equations JEE 2018?

Students write the rearranged equation as f'(x) sin x − f(x) cos x = +sin²x instead of the correct −sin²x. Integration then produces (f(x)/sin x)' = +1, so the wrong general solution f(x) = (x + K) sin x.

Applying the condition f(π/6) = −π/12 forces a specific negative K. The resulting value at x = π/4 differs from −π/(4√2). Students therefore accept statement A as true because the numbers no longer match. This sign error alone flips the entire answer key for them.

What are two other differential equations practice questions that test the same skills?

Question 1. Let f : (0, π) → R be twice differentiable such that

for all x ∈ (0, π). Given f(π/6) = π√3/12, find f(π/3).

Apply L'Hospital's rule to the 0/0 form to reach −f(x) sin x − f'(x) cos x = −cos²x. Rearrangement yields f'(x) cos x + f(x) sin x = cos²x. This numerator is exactly cos²x times the derivative of f(x)/cos x, so (f/cos x)' = 1. Integration gives f(x) = (x + C) cos x. The given value forces C = 0, hence f(x) = x cos x and f(π/3) = π/6.

Question 2. Suppose a limit condition on a twice differentiable f leads to (f(x)/x)' = 2 on (0, π) with f(π/4) = 0. Check whether f' has a root in (0, π) and whether f''(π/2) + f(π/2) > 0.

Integration yields f(x)/x = 2x + K, so f(x) = 2x² + Kx. Substitute the point to obtain K = −π/2, hence f(x) = 2x² − (π/2)x. Then f'(x) = 4x − π/2 = 0 at x = π/8 ∈ (0, π), confirming a root exists. Also f''(x) = 4 and f(π/2) = π²/4, so f''(π/2) + f(π/2) = 4 + π²/4 > 0.

Both questions reward the same exact recognition of the derivative quotient inside the limit and the same monotonicity or intermediate-value arguments used in the 2018 paper. You can search for these and similar problems in the past-paper archive by chapter to drill the pattern.

How can you prove sin x > x − x³/6 on (0, π) using only derivatives?

Define g(x) = sin x − x + x³/6. Then g(0) = 0 and g'(0) = 0. Compute g''(x) = x − sin x. On (0, π) the standard inequality x > sin x holds, so g''(x) > 0. A positive second derivative means g' is strictly increasing. Because g'(0) = 0 it follows that g'(x) > 0 for all x ∈ (0, π). Therefore g itself is strictly increasing from g(0) = 0, forcing g(x) > 0 on (0, π). This establishes sin x > x − x³/6 using only calculus and the intermediate value theorem on the derivatives.

What are the key takeaways for handling limit-defined differential equations in JEE Advanced?

Always rewrite the limit expression until it matches the numerator of a known derivative quotient. Never assume the constant before using the given point. Solve for it exactly as shown or the later numerical checks will fail.

For inequality options, construct an auxiliary function like g(x) above and check its derivatives at the boundaries. Verify the sign of the second derivative, then integrate the information upward.

Under exam pressure, write f(x) explicitly and test each statement independently rather than guessing combinations. This method converts any similar limit into a differential equation in under four minutes once the quotient pattern is drilled.

Solve the Limit, Continuity and Differentiability JEE 2020 Question Solution for extra practice on related limit tricks. If you get stuck on a similar problem during your preparation, photograph the doubt and get a step-by-step solution with diagrams where needed.